In triangle $ABC$ the medians $\overline{AD}$ and $\overline{CE}$ have lengths $18$ and $27$, respectively, and $AB=24$. Extend $\overline{CE}$ to intersect the circumcircle of $ABC$ at $F$. The area of triangle $AFB$ is $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$.

Explanation: [asy] size(150); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(8); pair A=(0,0), B=(24,0), E=(A+B)/2, C=IP(CR(A,3*70^.5),CR(E,27)), D=(B+C)/2, F=IP(circumcircle(A,B,C),E--C+2*(E-C)); D(D(MP("A",A))--D(MP("B",B))--D(MP("C",C,NW))--cycle); D(circumcircle(A,B,C)); D(MP("F",F)); D(A--D); D(C--F); D(A--F--B); D(MP("E",E,NE)); D(MP("D",D,NE)); MP("12",(A+E)/2,SE,f);MP("12",(B+E)/2,f); MP("27",(C+E)/2,SW,f); MP("18",(A+D)/2,SE,f); [/asy]
Applying Stewart's Theorem to medians $AD, CE$, we have:
\begin{align*} BC^2 + 4 \cdot 18^2 &= 2\left(24^2 + AC^2\right) \\ 24^2 + 4 \cdot 27^2 &= 2\left(AC^2 + BC^2\right)  \end{align*}
Substituting the first equation into the second and simplification yields $24^2 = 2\left(3AC^2 + 2 \cdot 24^2 - 4 \cdot 18^2\right)- 4 \cdot 27^2$ $\Longrightarrow AC = \sqrt{2^5 \cdot 3 + 2 \cdot 3^5 + 2^4 \cdot 3^3 - 2^7 \cdot 3} = 3\sqrt{70}$.
By the Power of a Point Theorem on $E$, we get $EF = \frac{12^2}{27} = \frac{16}{3}$. The Law of Cosines on $\triangle ACE$ gives
\begin{align*} \cos \angle AEC = \left(\frac{12^2 + 27^2 - 9 \cdot 70}{2 \cdot 12 \cdot 27}\right) = \frac{3}{8} \end{align*}
Hence $\sin \angle AEC = \sqrt{1 - \cos^2 \angle AEC} = \frac{\sqrt{55}}{8}$. Because $\triangle AEF, BEF$ have the same height and equal bases, they have the same area, and $[ABF] = 2[AEF] = 2 \cdot \frac 12 \cdot AE \cdot EF \sin \angle AEF = 12 \cdot \frac{16}{3} \cdot \frac{\sqrt{55}}{8} = 8\sqrt{55}$, and the answer is $8 + 55 = \boxed{63}$.